Mixing named and unnamed unpacking

[DavisVaughan]

library(dotty)

.[apple = x, banana] <- list(w = 1, x = 2, y = 3, z = 4)
apple
banana

Both of these return 2, is that right?

In general getting this "right" is probably quite hard.

Maybe it should be "if any are named, all must be named"?

[kevinushey]

Maybe it should be "if any are named, all must be named"?

I think that is probably the simplest way forward to reduce confusion. If not, then I think the right approach would be to implement argument matching in a way similar to R.

[DavisVaughan]

I vote for simplest and strictest now and if people revolt you can always relax it without breaking anything